POJ-1273-Drainage Ditches 朴素增广路

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 70588   Accepted: 27436 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.  Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.  Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.  Input The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch. Output For each case, output a single integer, the maximum rate at which water may emptied from the pond. Sample Input 5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10 Sample Output 50 Source USACO 93

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解题思路：

是一道裸的一般增广路算法的题目，就是计算水塘最大的流量排到附近的小河中。

增广路中可能含有反向弧，最后流量改进的时候对于反向弧的处理，既可以按照残留网络中的对应的正向弧那样增加流量也可以直接把这条反向弧的流量减少相应的流量。

因为我用这两种解法都能AC，说明效果是一样的。

源代码：

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

#define MAXN 210

struct Matrix
{
    int c,f;//容量，流量
};
Matrix Edge[MAXN][MAXN];//流及容量（邻接矩阵）
int M,N;//弧的条数，顶点个数
int s,t;//源点（1），汇点（n）
int residul[MAXN][MAXN];//残留网络
int qu[MAXN*MAXN],qs,qe;//队列、队列头、队列尾
int pre[MAXN];//pre[i]为增广路上顶点的访问标志
int vis[MAXN];//BFS算法中各个顶点的访问标志
int maxflow,min_augment;//最大流流量、每次增广时的可以改进的量
void find_augment_path()//BFS寻找增广路
{
    int i,cu;//cu为队列头顶点
    memset(vis,0,sizeof(vis));
    qs=0;qu[qs]=s;//s入队列
    pre[s]=s;vis[s]=1;qe=1;
    memset(residul,0,sizeof(residul));memset(pre,0,sizeof(pre));
    while(qs<qe&&pre[t]==0)
    {
        cu=qu[qs];
        for(i=1;i<=N;i++)
        {
            if(vis[i]==0)
            {
                if(Edge[cu][i].c-Edge[cu][i].f>0)
                {
                    residul[cu][i]=Edge[cu][i].c-Edge[cu][i].f;
                    pre[i]=cu;qu[qe++]=i;vis[i]=1;
                }
                else if(Edge[i][cu].f>0)
                {
                    residul[cu][i]=Edge[i][cu].f;
                    pre[i]=cu;qu[qe++]=i;vis[i]=1;
                }
            }
        }
        qs++;
    }
}
void augment_flow()//计算可改进量
{
    int i=t,j;//t为汇点
    if(pre[i]==0)
    {
        min_augment=0;return;
    }
    j=0x7fffffff;
    while(i!=s)//计算增广路上可改进量的最小值
    {
        if(residul[pre[i]][i]<j) j=residul[pre[i]][i];
        i=pre[i];
    }
    min_augment=j;
}
void update_flow()//调整流量
{
    int i=t;//t为汇点
    if(pre[i]==0)return;
    while(i!=s)
    {
        if(Edge[pre[i]][i].c-Edge[pre[i]][i].f>0)
            Edge[pre[i]][i].f+=min_augment;
        else if(Edge[i][pre[i]].f>0) Edge[pre[i]][i].f+=min_augment;//或者写成Edge[i][pre[i]]-=min_augment;
        i=pre[i];
    }
}
void solve()
{
    s=1;t=N;
    maxflow=0;
    while(1)
    {
        find_augment_path();//BFS寻找增广路
        augment_flow();//计算可改进量
        maxflow+=min_augment;
        if(min_augment>0) update_flow();
        else return;
    }
}
int main()
{
    int i;
    int u,v,c;
    while(scanf("%d %d",&M,&N)!=EOF)
    {
        memset(Edge,0,sizeof(Edge));
        for(i=0;i<M;i++)
        {
            scanf("%d %d %d",&u,&v,&c);
            Edge[u][v].c+=c;
        }
        solve();
        printf("%d\n",maxflow);
    }
    return 0;
}
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